proove that if a prime number is represented by this phormula p=(2^n)+1 with n>0 then n is a power of 2... it's all for you...
The important fact is that, we can factor:
$\displaystyle x^n+y^n$ where $\displaystyle n$ is odd.
Now, assume,
$\displaystyle 2^n+1$ is prime where $\displaystyle n$ is divisible by odd.
Then,
$\displaystyle 2^{(2k+1)j}+1$
Thus,
$\displaystyle (2^j)^{2k+1}+1$
Can be factored as,
$\displaystyle (2^j+1)(2^{2kj}-2^{(2k-1)j}+2^{(2k-2)j}-...+2^{2j}-2^j+1)$
It has a proper nontrivial factorization, thus it cannot be prime.
Thus, $\displaystyle n$ cannot be divisble by odd number that is, $\displaystyle 2^m$
Thus,
$\displaystyle 2^{2^m}+1$
Which were studied by Fermat (my favorite mathemation).